Identifying boundary of solid torus with itself by swapping coordinates Probably the following question is very elementary, but I couldn't find it anywhere.
Let $T$ be the solid 2-torus and let $\sim$ be the equivalence relation on $T$ generated by the relation $\{(\alpha,\beta) \sim (\beta,\alpha) \mid \alpha, \beta \in S^1\}$ on the boundary $\partial T=S^1\times S^1$. What kind of 3-fold is $T/{\sim}$?
I believe that it is $S^3$, but I can't prove it.
EDIT: My idea is the following: Take two copies of the 3-ball $B$. Represent $T=B\coprod B/{\sim}$ by identifying half of their respective boundaries (I'm not sure how exactly, maybe opposite quadrants..). Then the quotient $T/{\sim}$ should be given by identifying the other half of the boundaries, but it is known that this gives $S^3$. Can this work?
EDIT: This question has been answered here:
https://mathoverflow.net/questions/412288/quotient-of-solid-torus-by-swapping-coordinates-on-boundary
 A: I assume that you're familiar with the construction of compact surfaces by glueing edges of a square. The torus $T$ can be obtain from Fig. $1$ below by glueing edges of the same color (red, blue) with respect to given direction. Under relation $\sim$, every point is identified with its reflection over the diagonal (the dashed line). So the quotient space we want is given by Fig. $2$, here we glue the colored edges together.

It's easy to see that Fig. $2$ can be obtained from Fig. $3$ by deform the bottom edge into a point. To get the quotient spaces of Fig. 2, one can change the order of taking quotient; that is, to find the quotient space of Fig. 3 first, after that deform the image of the bottom edge (under the quotient map) into a point. But the quotient space $X$ corresponds to Fig. $3$ is just the Mobius strip. The black edges on the top and bottom of the square is joined to become $X$'s boundary, which is $S^1$. Moreover, two pairs of opposite vertices are identified; they become $2$ separated points $A$ and $B$ on $X$'s boundary and divide  the boundary into $2$ half-circles (each of them is the image of a black edge in the quotient). Hence the deformation of the bottom edge still make $X$ a Mobius strip, but with $A$ and $B$ identified. Therefore, $\partial T/{\sim}$ (edited $T\to \partial T$) is homeomorphic to a Mobius strip.
